One million pound-moles per day of a gas of the following composition is to be absorbed by n-heptane at -30°F and 550 psia in an absorber having 10 theoretical stages so as to absorb 50% of the...

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One million pound-moles per day of a gas of the following composition is to be absorbed by n-heptane at -30°F and 550 psia in an absorber having 10 theoretical stages so as to absorb 50% of the ethane. Calculate the required flow rate of absorbent and the distribution, in lbmol/h, of all the components between the exiting gas and liquidstreams.





Answered Same DayDec 31, 2021

Answer To: One million pound-moles per day of a gas of the following composition is to be absorbed by n-heptane...

David answered on Dec 31 2021
102 Votes
ID NO 376650-
FEED FLOW RATE - 1000000 LB MOL /HR
FEED COMPOSITION - K V
ALUE
AT -30 OF , P=550 PSIA
C1 94.9 2.85
C2 4.2 0.36
C3 0.7 0.016
NC4 0.1 0.017
NC5 0.1 0.004
PART A -
FOR C1 -COMPONENT –
XF = 0.5 , F =1000000 LB MOL/DAY =1000000/24 = 41666.67 LB MOL /HR
FORMULA -
Xbi = XFi /1+f (Ki -1)
XB = 0.949 /1+ 0.5 ( 2.85 -1) =0.949 /1+0.5 ( 1.85 ) = 0.949 / 1+0.925=0.949 /1.925=0.493
YB= Ki* xbi = 2.85 * 0.493 =1.405
FOR C2 COMPONENT -
Xbi = 0.042 /1+ 0.5 (0.36 -1 ) =0.042 / 1 + 0.5 * ( - 0.64)
Xbi = 0.042...
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